[tex]Dane:\\Q = 336 \ kJ = 336 \ 000 \ J\\T_1 = 20^{o}C\\T_2 = 60^{o}C\\\Delta T = T_2-T_1 = 60^{o}C - 20^{o}C = 40^{o}C\\c = 4200\frac{J}{kg\cdot^{o}C} \ - \ cieplo \ wlasciwe \ wody\\Szukane:\\m = ?\\\\Rozwiazanie\\\\Q = m\cdot c\cdot \Delta T \ \ \ |:(c\cdot \Delta T)\\\\m = \frac{Q}{c\cdot \Delta T}\\\\m = \frac{336000 \ J}{4200\frac{J}{kg\cdot^{o}C}\cdot40^{o}C}\\\\\boxed{m = 2 \ kg}[/tex]
